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3 years ago

Factor 2 from 19y+15x

Mathematics

1 answer:

Gnom [1K]3 years ago

8 0

Answer:

$2(9.5y + 7.5x)$

Step-by-step explanation:

We want to factor 2 from

$19y + 15x$

We rewrite the expresion to obtain;

$2 \times \frac{19}{2} y + 2 \times \frac{15}{2}x$

We now factor to get:

$2(\frac{19}{2} y + \frac{15}{2}x )$

$2 (9.5 y + 7.5x )$

Therefore we factor 2 out of the given expresion we get:

$2 (9.5y + 7.5x )$

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3 years ago

How many natural numbers from 78 to 234

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4 years ago

Read 2 more answers

Hamburger Hut sells regular hamburgers as well as a larger burger. Either type can include cheese, relish, lettuce, tomato, must

Studentka2010 [4]

Answer:

a) 40 different hamburgers can be ordered with exactly three extras

b) 20 different regular hamburgers can be ordered with exactly three extras

c) 7 different regular hamburgers can be ordered with at least five extras

Step-by-step explanation:

The order in which the extras are ordered is not important. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this problem:

2 options of hamburger(regular or larger)

6 options of extras(cheese, relish, lettuce, tomato, mustard, or catsup.).

(a) How many different hamburgers can be ordered with exactly three extras?

1 hamburger type, from a set of 2.

3 extras, from a set of 6. So

$C_{2,1}*C_{6,3} = \frac{2!}{1!(2-1)!}*\frac{6!}{3!(6-3)!} = 2*20 = 40$

40 different hamburgers can be ordered with exactly three extras

(b) How many different regular hamburgers can be ordered with exactly three extras?

3 extras, from a set of 6. So

$C_{6,3} = \frac{6!}{3!(6-3)!} = 20$

20 different regular hamburgers can be ordered with exactly three extras

(c) How many different regular hamburgers can be ordered with at least five extras?

Five extras:

5 extras, from a set of 6. So

$C_{6,5} = \frac{6!}{5!(6-5)!} = 6$

Six extras:

6 extras, from a set of 6. So

$C_{6,6} = \frac{6!}{6!(6-6)!} = 1$

6 + 1 = 7

7 different regular hamburgers can be ordered with at least five extras

8 0

3 years ago

It is said that happy and healthy workers are efficient and productive. A company that manufactures exercising machines wanted t

denpristay [2]

Answer:

Confidence interval is 0.7 ≤ p ≤ 0.8

The margin of error is 7.1 %

Step-by-step explanation:

We have to calculate a 98% confidence interval for the proportion.

The sample proportion is p=0.75.

The standard error of the proportion is:

Here we have, the proportion or point estimate given by

= 150/200 = 0.75

Sample size, n = 200

The formula for confidence interval, CI, given a proportion, is;

At 98% z = ±2.326348

Plugging in the values of, , z and n we get;

CI = 0.6787704 ≤ p ≤ 0.8212296

To one decimal place, we have

CI = 0.7 ≤ p ≤ 0.8

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3 years ago